What is the (e^{x^2-1})^' ?

The (e^{x^2-1})^' is e^{x^2-1}*2x

What is the first (e^{x^2-1})^' ?

The first (e^{x^2-1})^' is e^{x^2-1}*2x

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(e^{x^{2} - 1})^{^{'}}

e^{x^{2} - 1} \cdot 2 x

Solution steps

(e^{x^{2} - 1})^{'}

Apply the chain rule:

e^{x^{2} - 1} (x^{2} - 1)^{'}

= e^{x^{2} - 1} (x^{2} - 1)^{'}

(x^{2} - 1)^{'} = 2 x

= e^{x^{2} - 1} \cdot 2 x

\frac{\partial}{\partial x} (- 2 x y)

\frac{d}{d x} ((x^{3} - 4 x + 1) e^{x})

\frac{d}{d x} (arcsec^{2} (x))

x \to 0 lim (\frac{5}{tan ( 3 x )})

8 \frac{d y}{d x} + 7 e^{x + y} = 0